Orbital Mechanics II

Practice

practice problem 1

The following four statements about circular orbits are equivalent. Derive any one of them from first principles. This is a key relationship for a larger problem in orbital mechanics known as the virial theorem.

solution

Circular orbits arise whenever the gravitational force on a satellite equals the centripetal force needed to move it with uniform circular motion.

     
Fc  =  Fg
     
mv2  =  GMm
rp rp2
v2 =  Gm1
r

Substitute this expression into the formula for kinetic energy.

K =  1  m2v2
2
K =  1  m2 
Gm1
2 r
K =  1   Gm1m2
2 r

Note how similar this new formula is to the gravitational potential energy formula.

K = +  1   Gm1m2
2 r
Ug = −      Gm1m2
  r
K = −  1  Ug
2

The kinetic energy of a satellite in a circular orbit is half its gravitational energy and is positive instead of negative. When U and K are combined, their total is half the gravitational potential energy.

E = K + Ug  
 
E = −  1  Ug + Ug
2
E =  1  Ug
2
E = −  Gm1m2
2r

The gravitational field of a planet or star is like a well. The kinetic energy of a satellite in orbit or a person on the surface sets the limit as to how high they can "climb out of the pit". A satellite in a circular orbit is halfway out of the pit (or halfway in, for you pessimists).

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practice problem 2

Determine the minimum energy required to place a large (five metric ton) telecommunications satellite in a geostationary orbit.

solution

practice problem 3

solution

practice problem 4

solution